The shortest wavelength of the spectral lines in the Lyman series of hydrogen spectrum is 915 Å. The longest wavelength of spectral lines in the Balmer series will be _______ Å. Numerical answer type. Work it out, then open the answer.

Answer: 6588. Lyman shortest: 1/R = 915 Å Balmer longest ( 3 → 2): (1)/(λ) = 5R/36 ⇒ λ = 36/5R λ = 36/5× 915 = 6588 Å Answer: 6588 Å

The shortest wavelength of the spectral lines in the Lyman series of… | JEE Main 2024 (5 Apr, Shift 2)

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Hydrogen Spectrum & Spectral Series

The shortest wavelength of the spectral lines in the Lyman series of hydrogen spectrum is $915\text{ Å}$ . The longest wavelength of spectral lines in the Balmer series will be _______ $\text{Å}$ .

Numerical answer type. Work it out, then open the answer.

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Answer: 6588

Lyman shortest: $\frac{1}{R} = 915$ Å
Balmer longest ( $3 \to 2$ ): $\frac{1}{\lambda} = \frac{5R}{36} \Rightarrow \lambda = \frac{36}{5R}$
$\lambda = \frac{36}{5}\times 915 = 6588$ Å
Answer: 6588 Å

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